How do I prove this transcendental equation has a solution? I am trying to prove that for the following equation, there is a B that solves it (c is a constant):
$1-B = e^{-cB}$
I understand this is a transcendental equation, but how do I prove there is a B that solves it?
I need a non-zero solution. c > 1
 A: Edit: The question was changed to eliminate the solution $B=0$.
First answer: By inspection, $B=0$ is a solution. In the case $c=1$, it is the only solution. 
Answer to modified question: Consider the function $f(x)=1-x-e^{-cx}$. We have $f'(x)=ce^{-cx}-1$. This is positive at $x=0$, since $c\gt 1$. Thus  by continuity $f'(x)\gt 0$ for a while after $x=0$. 
It follows that $f(x)$ is increasing for a while past $x=0$, and in particular is positive for some $a\gt 0$.
When $x$ is large enough, $f(x)$ is negative, since $\lim_{x\to\infty}e^{-cx}=0$.
Thus by the Intermediate Value Theorem, there is an $x\gt 0$ such that $f(x)=0$.
Remark: By closer examination of the derivative, one can show that there is exactly one $x\gt 0$ such that $f(x)=0$. 
A: Note that for $c \gt 1$, for $B$ slightly greater than zero $1-B \gt e^{-cB}$ (you can use the derivatives to show this).  At $B=1$ we have $1-B \lt e^{-cB}$ and there must be a point where they cross.
A: I'm going to use $x$ for $B$:
$1-x = e^{-cx}$
http://www.wolframalpha.com/input/?i=1-x+%3D+e%5E%28-cx%29 solves for x as


*

*1 $$x = \frac {W(-c*e^{-c} )+c} {c}$$ where W is the product log function. 


To prove this, we need to show that $$(cx-c)e^{cx-c}+c*e^{-c}=0$$
Let's use an alternate form of the original equality(this is trivial):


*

*2 $$(x-1)e^{cx}=-1$$


Assume $(cx-c)e^{cx-c}+c*e^{-c}=y$ 
$$(cx-c)e^{cx-c}=y-c*e^{-c}$$
Rewrite this as $$\frac {(cx-c)e^{cx}} {e^c}=y-c*e^{-c}$$
Now, from 2 we have $$(cx-c)e^{cx}=-c$$
Substituting that gives $$-\frac {c} {e^c}=y-c*e^{-c}$$
which trivially leads to $y=0$, which then leads to 1 above, using the definition  of $W$.
